David Levy

If I understand it correctly, the aim of this article is to prove that Sgr A*.is the real S2 center of mass. However, it seems to me that 2002 data contradicts this requirement: In 2002 S2 was at the closest position to Sgr A*. (nearly coincident). "In 2002, S2 was positionally nearly coincident with Sgr A* and thus confused with the NIR counterpart of the MBH." Actually this kind of nearly coincident between S2 and Sgr A* could negativly effect the requirment for center of mass in Keplerian ellips. In any case, based on the pure data from 2002 it seems that there is no fit: "From this analysis, it is clear that the weight of the 2002 data will influence the resulting orbit fits, since these points will systematically change the orbit figure at its pericenter. At the same time we have no plausible explanation for the increase in brightness and the systematic residuals in the 2002 data; in particular a confusion event seems unlikely. Thus, it is clear that using the 2002 data will affect the results, but we cannot decide whether it biases towards the correct solution or away from it. Therefore we use in the following two options: a) we include the 2002 data with the increased error bars; b) we completely disregard the 2002 data of S2." Therefore, it was decided to fix the data of 2002 as follow: Fig. 10.— The 2002 data of S2. The grey symbols show the measured positions, the errors are as obtained from the standard analysis and are not yet enlarged by the procedure described in section 3.5. The black dots are the positions predicted for the observation dates using an orbit fit obtained from all data other than 2002. The blue shaded areas indicate the uncertainties in the predicted positions resulting from the uncertainties of the orbital elements and of the potential, taking into account parameter correlations. After fixing the data, they got better fit: "From the numbers it seems that the fit excluding the 2002 data agrees better with the expectations for the coordinate system (equation 4) than the fit including it." It is also stated: "Systematic problems of the coordinate system could be absorbed into the orbital fitting by allowing the center of mass to have an offset from 0/0 and a non-zero velocity, at the cost of not being able to test the coincidence of the center of mass with radio Sgr A*. From this analysis, it is clear that the weight of the 2002 data will influence the resulting orbit fits, since these points will systematically change the orbit figure at its pericenter. I wonder how could it be that the center of mass has an offset and a non zero velocity. Those are key requirements from any center of mass. If it has velocity and offset – how can we use it as a center of mass?

Thanks It is stated: Draft version October 26, 2008 Is there any more updated data?

Thanks Janus Your explanation is very interesting. It's the first time that I get such explanation about gravity. So it acts as a sinusoidal gravity wave going up and down as it cross different mass density zones. Do we have any other example which could confirm this idea? I had the impression that there could be some sort of tidal but not that kind of gravity change. In any case, those different densities must be quite neglected comparing to Sgr A* mass (less than 1%). "the strongest constraint that can be placed on these remnants is that their total mass comprises less than one percent of the mass of the supermassive black hole." Therefore, how could it be that there is so severe impact on S2 Keplerian ellips? Please look again on the following Diagram. http://www.universetoday.com/wp-content/uploads/2010/08/nature01121-f2.22.jpg We can see clearly that Sgr A* is not located at a symmetric point in that ellipse. (It is located at the bottom left side) However, in your diagram the red circle - which represents Sgr A* , is fully symmetric. Hence, how can we use an asymmetric point as a center of the ellipse? The Sun is located perfectly at a symmetric point of the moon orbit ellipse.

Thanks That is clear https://imagine.gsfc.nasa.gov/features/yba/CygX1_mass/gravity/sun_mass.html So,we can calculate Sun mass directly from Earth/Sun orbit without any info about Earth or moon mass. However, it must be perfect Keplerian ellipse. How that impact S2?

O.K. Let's focus on Moon /Earth / Sun Example In our example we have only eliminated the light on Earth. However, it is still there. Its gravity has a severe impact on Moon / Sun orbit. Therefore, it is clear to me that we can't use this formula for this example as it is stated: "This means that influences from any third body are neglected",

It is stated: "The gravitational two-body problem concerns the motion of two point particles that interact only with each other, due to gravity. This means that influences from any third body are neglected." So, if S2 orbit was a perfect Keplerian ellips, than yes, we could use this formula. However, S2 orbit isn't a perfect one. It is also stated clearly that there are some other gravity forces on S2: https://en.wikipedia.../wiki/S2_(star) "The motion of S2 is also useful for detecting the presence of other objects near to Sagittarius A*. It is believed that there are thousands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs) distributed in the volume through which S2 moves. Hence, it seems to me that this formula is not relevant to S2 case. How can we claim in one hand that "This means that influences from any third body are neglected", While on the other hand it is stated that: "housands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs) distributed in the volume through which S2 moves"? How can it work together? Therefore, I would consider it as a mistake.

Very simple: Moon Mass - 7.342×1022 kg Sun Mass – 1.98855×1030 kg Earth mass = 5.97219 * 10 24 kg Distance : 149,600,000 Km Please use the following formula to calculate the gravity force between Earth/Sun. https://www.easycalculation.com/physics/classical-physics/newtons-law.php I have got: F= 3.5922175 ×1022 N Now change it to calculate M1 mass based on M2. However, instead of using the mass of the Earth, please use Moon mass. The outcome is that the Sun mass is: 1.6409299575189575e+24 That is a severe mistake!!! Therefore, we can't calculate the real value of the Sun without verifying first the Earth mass.

It seems to me that you assume that the Earth mass is well known. However, in this example we have no info about the earth mass. Hence, how can we calculate the sun mass without any info about Earth or moon mass?

O.K. So we agree that the Earth is invisible. We can only see the Moon and the Sun. We know the total mass of the Moon, but we need to verify the mass of the Sun. We also know that moon's orbit around the Sun isn't a perfect Keplerian ellips 1. No, you can't calculate the Sun mass directly from the moon orbital period without extract the Earth Mass first 2. - That is correct. 3 - Can you please elaborate how can you do it? How the departure of the moon's orbit from a perfect ellipse can give you indication about Earth mass? With Regards to S2. You claim correctly: "2. From the shape of the moon's orbit (not exactly an ellipse) we would know that there was another mass present". However, S2 has no perfect Keplerian ellips, than it is clear indication that there is another mass present. So, why do we ignore that vital info? How can we do it? If we can only see the moon and the Sun, how can we extract the Sun mass without the information about the moon mass? (Please remember - there is no info at all about the Earth!)

O.K. Gravity is gravity. It works for all and in the same way. if you are moon, Earth, Sun, S2 or even BH, you have to obey to gravity law. In order to understand this statement, let's start with the following explanation about Earth/Moon Orbit and their centre of mass. http://www.animations.physics.unsw.edu.au/jw/gravity.htm Orbits and the centre of mass It is a common simplification to say that 'the moon's orbit is a circle about the earth': the gravitational attraction of the moon towards the earth provides the centripetal force for the moon's orbit. From Newton's third law, of course, the moon attracts the earth with a force of equal magnitude, which accelerates the earth. So the earth and the moon each trace (almost perfectly) circular orbits about their common centre of mass. The sketched below illustrates this, but is not to scale. As you can see, the earth and the moon each trace (almost perfectly) circular orbits about their common centre of mass. This center of mass orbits the Sun. Hence, the Earth and the Moon orbits their common center of mass, while this virtual point of center mass orbits the Sun. So, this virtual common center of Earth/Moon mass point orbits the Sun in a perfect Keplerian ellipse. (Although some people would assume that the Earth orbits the sun in a perfect Keplerian ellipse - and that is a mistake), However, the Earth is significantly heavier than the Moon, therefore, this Common center of mass is located in its radius. Therefore, it isn't so big mistake to assume that the Earth orbits the sun in a almost perfect Keplerian ellipse. Never the less, the moon itself doesn't orbit the Sun in a perfect Keplerian ellips. The only way to verify this issue is by monitoring its orbital path. So, assuming that we can't see the Earth, by monitoring the Noon orbital path we can easily verify that it doesn't orbit the Sun in a perfect Keplerian ellips. If we ignore that vital information, we could assume that the total mass of that Earth/Moon common center of mass is as big as the moon itself. So we have neglected completely the Earth mass.Therefore, if we will try to extract the Sun mass directly out of the Moon mass and its none perfect Keplerian ellips, we would make a severe mistake which is proportional to the Earth/moon mass ration. That is not an error of 1%, but a huge error!!! Don't forget that the distance between the Moon/Earth is about 1.5 Sec of light and the Earth/Sun is above 8 Minutes of light. Therefore, just by looking on the Moon/Sun orbit it might be almost an impossible mission to verify that the moon orbit around the Sun isn't a perfect Keplerian ellips. However, there is one more issue which can help us in this verification. It is the moon speed. As it orbits the Common center of mass, we should easily verify that there are picks in its speed. If we can detect those picks, we could calculate the time duration that it takes the moon to complete one cycle around that virtual Earth/Moon common of mass. If we can verify the radius of that cycle, we can extract the value of Earth mass. With this info, we can easily get the real mass of the Sun. So that was all about our solar system. In the same way we have to solve S2 Orbital path around the SMBH. As its orbit isn't a perfect Keplerian ellips it is clear that there must be at least one level of common center of mass. First we must monitor S2 speed and see if there are picks in that speed. Each pick indicates that S2 had completed one cycle around its common center of mass. If we can verify the radius of that orbit we can extract the total mass of the object which it orbits. Then, we have to look carefully on that center of mass and verify if it orbits the SMBH is perfect Keplerian ellips. If it does, than by simple calculation of S2 and the other object mass, we can calculate the real mass of SMBH. Is it clear by now?

Please look at the following S2 orbit diagram http://www.universetoday.com/wp-content/uploads/2010/08/nature01121-f2.22.jpg It is quite clear that S2 is moving in and out the from Keplerian ellipse However, the time resolution is quite poor. (There is about one spot per year). In the same token please look at the following moon orbit around the Sun: https://en.wikipedia.org/wiki/Talk%3AOrbit_of_the_Moon#/media/File:Earth_orbit.svg We also see that it moves in and out from the Keplerian ellipse. So, if we could get better time resolution of S2 orbit diagram, we might get better understanding about its single point of mass.

For better understanding let's use the solar system as an example. Especially - Moon, Earth and Sun. So, let's isolate that system. However, in this example we will shut down Earth light. Hence, all we can see is Moon and Sun. We know the total mass of the Moon, but we are requested to calculate the Sun mass. We get full information about the moon orbit. We see that the moon orbits the Sun in one year, however, its movement deviates from a perfect Keplerian ellipse. It orbits the Sun with minor changes – a little bit up, a little bit down, sometimes it moves fast sometimes slow. Technically, we can ignore that vital data and could get into simple calculation that the total mass of the sun is as big as Earth mass. Is it correct? We can easily claim: So, yes, the moon orbit the Sun in one year. Could it be that the moon orbits the Sun with only 1% deviation from a perfect ellipse? Could it be that the Earth mass is deviated by only 1% from the Sun mass? Hence, we first must find the temporary location of the Moon's single point mass (which is – Earth), based on perfect Keplerian ellipse. It is quite difficult as we only see a moon deviation from perfect Keplerian ellipse. However, based on our ultra high powerful commuters, we should verify the temporary location of that Moon's single point mass (Earth), and even calculate its expected mass. Then we should try to verify if that Moon's single point mass (Earth) is orbiting the Sun in perfect Keplerian ellipse. If the answer is positive, then we can easily calculate the real mass of the Sun. Conclusion – There is no direct way to calculate a mass of an object which isn't in a perfect Keplerian ellipse orbit. Any deviation from Keplerian ellipse is critical. With regards to S2 We must first calculate the temporary location of S2 single point mass and its total mass. Than we have to verify if that S2 single point mass orbits the SAG A* in a perfect Keplerian ellipse. If so, we can easily extract the real mass of SAG A*.

How could it be that S2 orbit deviates from Keplerian ellipse and at what percentage? In the following article it is stated that (It is believed that) it is due to stars, dark stellar...: https://en.wikipedia.org/wiki/S2_(star) "The motion of S2 is also useful for detecting the presence of other objects near to Sagittarius A*. It is believed that there are thousands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs) distributed in the volume through which S2 moves. These objects will perturb S2's orbit, causing it to deviate gradually from the Keplerian ellipse that characterizes motion around a single point mass.[9] So far, the strongest constraint that can be placed on these remnants is that their total mass comprises less than one percent of the mass of the supermassive black hole.[10] However, it is clear (to me) that stars orbit must obey to Keplerian ellipse . I assume that this statement is due to our believe that S2 single point mass must be located at the SAG A*. Hence, could it be that it deviates from Keplerian ellipse as we believe that it must orbit SAG A*. In other words, if we eliminate this requirement, and verify S2 orbit based on pure Keplerian ellipse could it be that we might find that its single point mass is located at a different location than SAG A*? Did we try it? This information is very critical. In one hand we claim that S2 orbit deviates from Keplerian ellipse. So, we are well aware that there are some other forces which might have an impact on S2 orbit However, on the other hand we use a none Keplerian ellipse orbit to estimate SAG A* mass. https://en.wikipedia.org/wiki/Sagittarius_A* From the motion of star S2, the object's mass can be estimated as 4.1 million solar masses.[3] (The corresponding Schwarzschild radius is 0.08 AU/12 million km/7.4 million miles; 17 times bigger than the radius of the Sun.) How can we set SAG A* mass estimation on an orbit which is effected by (thousands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs)? Never the less, it is also stated that: "the strongest constraint that can be placed on these remnants is that their total mass comprises less than one percent of the mass of the supermassive black hole" So, does it mean that S2 orbit deviates from Keplerian ellipse by only 1%? Hence, can we understand that by using pure Keplerian ellipse S2 single point mass deviates from SAG A* location by only 1%? If so, how far the real S2 single point mass is located with regards to SAG A* location?

Well, it's quite disappointing that the Accretion disk radius is so mysterious (for all of us at any language) . I hope to find it later on.

Thanks Hence: The Radius of Sagittarius_A* Event horizon = 12 Million Km. However, that exactly the radius of its SMBH mass. Is it O.K.? I had the impression that the location of event horizon should be quite outwards than the SMBH radius mass. It is just on the edge. So, some SMBH mass at the edge could technically drift outside the event horizon. In this case, could it be that some light from the SMBH mass might escape from the event horizon?

What is the radius of Sagittarius_A* Event horizon? I couldn't find a direct number, however based on the following info its quite easy to calculate. https://en.wikipedia.org/wiki/Sagittarius_A* First noticed as something unusual in images of the center of the Milky Way in 2002,[34] the gas cloud G2, which has a mass about three times that of Earth, was confirmed to be likely on a course taking it into the accretion zone of Sgr A* in a paper published in Nature in 2012.[35] Predictions of its orbit suggested it would make its closest approach to the black hole (a perinigricon) in early 2014, when the cloud was at a distance of just over 3000 times the radius of the event horizon (or ≈260 AU, 36 light-hours) from the black hole. Hence: Radius of Sagittarius_A* Event horizon = 36 light-hours / 3000 36 light-hours = 36 * 3.88531026 e10 kilometers Radius of Sagittarius_A* Event horizon = 46.6237 * e7 kilometers = 460 Million Km. However, the accretion disk must be located outwards the radius of Sagittarius_A* Event horizon. Therefore, this info should give us an indication for the minimal radius of the accretion disk. Is it correct?

Thanks

Can you please give me your best understanding/estimation for the radius of the accretion disk. Could it be in the range of 100 Million Km?

Thanks So, the radius of the SMBH is 12 Million Km, while the diameter of the accretion disk is: 37 /12 * 12 Million Km = 37 Million Km Hence, the radius of the accretion disk should be 37/2 Million Km = 18.5 Million Km. Do you agree with that?

Thanks for the article. https://arxiv.org/abs/1405.1456 Please advice if my estimation is correct: It is stated: "The best estimate for the Schwarzschild radius based on observational limits on the black hole mass and distance is 1 RS = 10.2 ± 0.5µas plus systematic errors (Genzel et al. 2010; Falcke & Markoff 2013); we adopt a value of 10 µas for this paper." So, the angular diameter of the SMBH is 1RS, which is estimated as: 1 RS = 10 µas However, in the following article it is stated that it is 37 μas. https://en.wikipedia.org/wiki/Sagittarius_A* Astronomers have been unable to observe Sgr A* in the optical spectrum because of the effect of 25 magnitudes of extinction by dust and gas between the source and Earth.[10] Several teams of researchers have attempted to image Sagittarius A* in the radio spectrum using very-long-baseline interferometry (VLBI).[11] The current highest-resolution measurement, made at a wavelength of 1.3 mm, indicated an angular diameter for the source of 37 μas.[12] At a distance of 26,000 light-years, this yields a diameter of 44 million kilometers. For comparison, Earth is 150 million kilometers from the Sun, and Mercury is 46 million kilometers from the Sun at perihelion. The proper motion of Sgr A* is approximately −2.70 mas per year for the right ascension and −5.6 mas per year for the declination.[13] In any case, let's assume that the diameter of the SMBH is 44 million kilometers (which is almost as the distance from mercury to the Sun). I couldn't find solid data on the accretion disk. It is just stated: " In support of this extrapolation, we note that Meyer et al. (2007) do model accretion disk structure on a scale of a few RS" So, if we assume that 5 might represent "a few", could it be that the diameter of the accretion disk equal to: 5 * 44 Million Km = (about) 200 Million km. Would you consider it as a reasonable estimation?

Quote We should care. It is one of the most important data on the accretion disk. Please try to offer even an estimation.

With regards to the accretion disk at the Milky Way; Please advice the minimal and the maximal radius of that ring.

Is it a quiz? I have no clue. So, can it hold them together as some sort of a pack...? What do you mean by "very important role"? How does the magnetic force effect the particles? Does it have any effect on the particles rotation around the SMBH? If yes, how?

Sorry if I didn't understand your message correctly. I apologise now. So, do you mean that the message should be? "There is magnetic force between the particles but this force has no effect on the particles, therefore it can't hold them together as a pack". In this case, can you please advice what could be contribution of that magnetic force on the particles? Shall we just ignore it?

There are several types of plasma. At the accretion disk there is Astrophysical plasmas. https://en.wikipedia.org/wiki/Plasma_(physics)#Fluid_model Astrophysical plasmas are also observed in Accretion disks around stars or compact objects like white dwarfs, neutron stars, or black holes in close binary star systems.[30] Plasma is associated with ejection of material in astrophysical jets, which have been observed with accreting black holes[31] or in active galaxies like M87's jet that possibly extends out to 5,000 light-years. https://en.wikipedia.org/wiki/Astrophysical_plasma An astrophysical plasma is a plasma (a highly ionized gas) whose physical properties are studied as part of astrophysics. Much of the baryonic matter of the universe is thought to consist of plasma,[2] a state of matter in which atoms and molecules are so hot, that they have ionized by breaking up into their constituent parts, negatively charged electrons and positively charged ions. Because the particles are charged, they are strongly influenced by electromagnetic forces, that is, by magnetic and electric fields.[citation needed] All astrophysical plasmas are likely influenced by magnetic fields. So, all astrophysical plasmas are likely influenced by magnetic fields. Now let's read again the meaning of magnetic field in that kind of plasma: http://www.einstein-...ights/accretion "The most efficient mechanism to re-distribute angular momentum involves plasma matter, in which the different particles influence each other via weak magnetic fields. It is stated clearly : " the different particles influence each other via weak magnetic fields" Hence, different particles influence each other via weak magnetic fields. However, magnetic field is defined in terms of force on moving charge in the Lorentz force law. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfie.html#c1 "The magnetic field B is defined in terms of force on moving charge in the Lorentz force law." Hence, what is the meaning of that magnetic force? Why do you claim that there is no magnetic force between the nearby particles in the accretion plasma?